Problem: You have found the following ages (in years) of all 4 tigers at your local zoo: $ 5,\enspace 6,\enspace 13,\enspace 14$ What is the average age of the tigers at your zoo? What is the variance? You may round your answers to the nearest tenth.
Solution: Because we have data for all 4 tigers at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $4$ ages and divide by $4$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{5 + 6 + 13 + 14}{{4}} = {9.5\text{ years old}} $ Find the squared deviations from the mean for each tiger. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $5$ years $-4.5$ years $20.25$ years $^2$ $6$ years $-3.5$ years $12.25$ years $^2$ $13$ years $3.5$ years $12.25$ years $^2$ $14$ years $4.5$ years $20.25$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{20.25} + {12.25} + {12.25} + {20.25}} {{4}} $ $ {\sigma^2} = \dfrac{{65}}{{4}} = {16.25\text{ years}^2} $ The average tiger at the zoo is 9.5 years old. The population variance is 16.25 years $^2$.